Question

How to solve without the l'Hospital's rule? `lim_(x->0) (xcos^2(x))/(x+tan(3x))`

Answer 1

`1/4`

Explanation:

`"You could use the Taylor series expansion."`
`cos(x) = 1 - x^2/2! + x^4/4! - ...`
`tan(x) = x + x^3/3 + 2 x^5/15 + ...`
`=> cos^2(x) = 1 - x^2 + x^4 (1/4 + 2/24) ...`
`= 1 - x^2 + x^4/3 ...`
`=> tan(3x) = 3x + 9 x^3 + ...`

`=> (x*cos^2(x))/(x+tan(3x)) =`
`(x - x^3 + x^5/3 ...)/(4x + 9 x^3 + ...)`
`x->0 => " higher powers disappear"`
`= (x - ...)/(4x + ...)`
`= 1/4`

Answer 2

Please see below.

Explanation:

`(xcos^2x)/(x+tan3x) = 1/(1+(sin3x)/(xcos3x))*cos^2x`

`= 1/(1+3((sin3x)/(3x)) * 1/(cos3x))*cos^2x`

Note that `lim_(xrarr0)(sin3x)/(3x) = 1` and `lim_(xrarr0)cosx = 1`

So in the limit, we have:

`1/(1+3(1) * 1/1) * (1)^2 = 1/(1+3) = 1/4`